MCQ
$6.24 \times 10^{19}$ electrons is equal approximately to:
- ✓$10$ coulombs.
- B$96500$ coulombs.
- Cone electron volt.
- D$0.1F.$
✓
Answer
Correct option: A.
$10$ coulombs.
Since $ 6.24 \times 10^{18}$ electrons $= 1$ coulomb
$\therefore 6.24 \times 10^{19}$ electrons $= 10$ coulomb
$\Big(\frac{6.24\times10^{19}}{6.24\times10^{18}}\Big)$
$\therefore 6.24 \times 10^{19}$ electrons $= 10$ coulomb
$\Big(\frac{6.24\times10^{19}}{6.24\times10^{18}}\Big)$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Insulin production and its action in human body are responsible for the level of diabetes. This compound belongs to which of the following categories
Ammonia is a Lewis base. It forms complexes with cations. Which one of the following cations does not form complex with ammonia
The $IR$ stretchin frequencies of free $CO,$ and $CO$ in $[V(CO)_6]^-, [Cr(CO)_6]^-$ and $[Mn(CO)_6]^-$ are $2143\,cm^{-1} , 1860\,cm^{-1} , 2000 \,cm^{-1}$ and $2090 \,cm^{-1}$ , respectively. Then correct statement about metal carbonyls is
→ $\xrightarrow{{PC{l_5}}}A\xrightarrow{{LiAl{H_4}}}B$$\xrightarrow[{PCC}]{}C\xrightarrow[\Delta ]{{\overline {OH} }}D$
→Compound $D$ is :-
The element which forms oxides in all oxidation states $ + I$ to $ + V$ is
→Primary and secondary amines can be distinguished by
What should be the state of compounds which decompose on passing electric current?
In the following reaction $ 'A'$ is
→$\begin{array}{*{20}{c}}
{{C_2}{H_5}MgBr + {H_2}C - C{H_2}\xrightarrow{{{H_2}O}}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\backslash \,\,\,\,/} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O}
\end{array}A$
All of the following statements apply to proteins except
The following complexes are given :
→$(1)$ trans $-[Co(NH_3)_4 Cl_2]^+$
$(2)$ cis $-[Co(NH_3)_2 (en)_2]^{3+}$
$(3)$ trans $-[Co(NH_3)_2(en)_2]^{3+}$
$(4)$ $NiCl^{2-}_4$
$(5)$ $TiF^{2-}_6$
$(6)\, CoF^{3-}_6$
Choose the correct code